Legendre Polynomials

In many problems, the real number $p,$ appearing in the Legendre Equation (9.4.1), is a non-negative integer. Suppose $p = n$ is a non-negative integer. Recall

$$a_{k+2} = - \frac{(n-k)(n+k+1)}{(k+1)(k+2)} a_k, \; k=0,1,2,\ldots.$$ (9.4.5)

Therefore, when $k = n,$ we get

$$a_{n+2} = a_{n+4} = \cdots = a_{n+2m} = \cdots = 0 \; {\mbox{ for all positive integer}} \; m.$$

Case 1: Let $n$ be a positive even integer. Then $y_1$ in Equation (9.4.2) is a polynomial of degree $n.$ In fact, $y_1$ is an even polynomial in the sense that the terms of $y_1$ are even powers of $x$ and hence $y_1(-x) = y_1(x).$
Case 2: Now, let $n$ be a positive odd integer. Then $y_2(x)$ in Equation (9.4.3) is a polynomial of degree $n.$ In this case, $y_2$ is an odd polynomial in the sense that the terms of $y_2$ are odd powers of $x$ and hence $y_2(-x) = - y_2(x).$

In either case, we have a polynomial solution for Equation (9.4.1).

DEFINITION 9.4.3   A polynomial solution $P_n(x)$ of (9.4.1) is called a LEGENDRE POLYNOMIAL whenever $P_n(1) = 1.$

Fix a positive integer $n$ and consider $P_n(x) = a_0 + a_1 x + \cdots + a_n x^n.$ Then it can be checked that $P_n(1) = 1$ if we choose

$$a_n = \frac{(2n)!}{2^n (n!)^2} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}.$$

Using the recurrence relation, we have

$$a_{n-2} = -\frac{(n-1)n}{2(2n-1)} a_n = -\frac{(2n-2)!}{2^n (n-1)! (n-2)!}$$

by the choice of $a_n.$ In general, if $n - 2m \geq 0,$ then

$$a_{n-2m} = (-1)^m \frac{(2n-2m)!}{2^n m! (n-m)! (n-2m)!}.$$

Hence,

$$\sum_{m=0}^M (-1)^m \frac{(2n-2m)!}{2^n m! (n-m)! (n-2m)!} x^{n-2m},$$ (9.4.6)

where $M = \displaystyle\frac{n}{2}$ when $n$ is even and $M = \displaystyle\frac{n-1}{2}$ when $n$ is odd.

PROPOSITION 9.4.4   Let $p = n$ be a non-negative even integer. Then any polynomial solution $y$ of (9.4.1) which has only even powers of $x$ is a multiple of $P_n(x).$

Similarly, if $p = n$ is a non-negative odd integer, then any polynomial solution $y$ of (9.4.1) which has only odd powers of $x$ is a multiple of $P_n(x).$

Proof. Suppose that $n$ is a non-negative even integer. Let $y$ be a polynomial solution of (9.4.1). By (9.4.4)

$$y = c_1 y_1 + c_2 y_2,$$

where $y_1$ is a polynomial of degree $n$ (with even powers of $x$ ) and $y_2$ is a power series solution with odd powers only. Since $y$ is a polynomial, we have $c_2 = 0$ or $y = c_1 y_1$ with $c_1 \neq 0.$
Similarly, $P_n(x) = c_1^{\prime} y_1$ with $c_1^{\prime} \neq 0.$ which implies that $y$ is a multiple of $P_n(x).$ A similar proof holds when $n$ is an odd positive integer. height6pt width 6pt depth 0pt

We have an alternate way of evaluating $P_n(x).$ They are used later for the orthogonality properties of the Legendre polynomials, $P_n(x)$ 's.

THEOREM 9.4.5 (Rodrigu$\dot{e}$ s Formula)   The Legendre polynomials $P_n(x)$ for $n = 1, 2, \ldots,$ are given by

$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n.$$ (9.4.7)

Proof. Let $V(x) = (x^2 - 1)^n.$ Then $\frac{d}{dx} V(x) = 2 n x (x^2 -1)^{n-1}$ or

$$(x^2 - 1) \frac{d}{dx} V(x) = 2 n x (x^2 -1)^{n} = 2n x V(x).$$

Now differentiating $(n+1)$ times (by the use of the Leibniz rule for differentiation), we get

$$(x^2 - 1) \frac{d^{n+2}}{dx^{n+2}} V(x) + 2 (n+1) x \frac{d^{n+1}}{dx^{n+1}} V(x) + \frac{2n (n+1)}{1\cdot 2} \frac{d^{n}}{dx^{n}} V(x)$$

$$- 2nx \frac{d^{n+1}}{dx^{n+1}} V(x) - 2n(n+1) \frac{d^{n}}{dx^{n}} V(x) = 0.$$

By denoting, $U(x) = \frac{d^{n}}{dx^{n}} V(x),$ we have

$$(x^2 - 1)U^{\prime\prime} + U^\prime \{ 2(n+1) x - 2n x\} + U \{n(n+1) - 2 n (n+1) \} =$$ 0

$${\mbox{or}} \;\;\; \;\; (1 - x^2)U^{\prime\prime} - 2 x U^\prime + n(n+1) U = 0.$$

This tells us that $U(x)$ is a solution of the Legendre Equation (9.4.1). So, by Proposition 9.4.4, we have

$$P_n(x) = \alpha U(x) = \alpha \frac{d^{n}}{dx^{n}} (x^2 - 1)^n \;\; {\mbox{ for some }} \;\; \alpha \in {\mathbb{R}}.$$

Also, let us note that

$$\frac{d^{n}}{dx^{n}} (x^2 - 1)^n = \frac{d^{n}}{dx^{n}} \{(x - 1)(x+1)\}^n$$

$$= n! (x+1)^n + \; {\mbox{ terms containing a factor of }} \; (x-1).$$

Therefore,

$$\frac{d^{n}}{dx^{n}} (x^2 - 1)^n\biggr\vert _{x=1} = 2^n n! \;\; {\mbox{ or, equivalently }}$$

$$\frac{1}{2^n n!} \frac{d^{n}}{dx^{n}} (x^2 - 1)^n\biggr\vert _{x=1} = 1$$

and thus

$$P_n(x) = \displaystyle\frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n.$$

height6pt width 6pt depth 0pt

EXAMPLE 9.4.6  

1. When $n = 0, \; P_0(x) = 1.$
2. When $n = 1, \; P_1(x) = \displaystyle\frac{1}{2} \frac{d}{dx} (x^2 - 1) = x.$
3. When $n = 2, \; P_2(x) = \displaystyle\frac{1}{2^2 2!} \frac{d^2}{dx^2} (x^2 - 1)^2 = \frac{1}{8} \{ 12 x^2 - 4\} = \frac{3}{2} x^2 - \frac{1}{2}.$

One may observe that the Rodrigu$\dot{e}$ s formula is very useful in the computation of $P_n(x)$ for ``small" values of $n.$

THEOREM 9.4.7   Let $P_n(x)$ denote, as usual, the Legendre Polynomial of degree $n.$ Then

$$\int_{-1}^1 P_n(x) P_m(x) \; dx = 0 \; {\mbox{ if }} \; m \neq n.$$ (9.4.8)

Proof. We know that the polynomials $P_n(x)$ and $P_m(x)$ satisfy

$$\bigl( (1 - x^2) P_n^\prime(x) \bigr)^\prime + n(n+1) P_n(x) = 0\; {\mbox{ and }}$$ (9.4.9)

$$\; \bigl( (1 - x^2) P_m^\prime(x) \bigr)^\prime + m(m+1) P_m(x) = 0.$$ (9.4.10)

Multiplying Equation (9.4.9) by $P_m(x)$ and Equation (9.4.10) by $P_n(x)$ and subtracting, we get

$$\bigl( n(n+1) - m(m+1) \bigr) P_n(x) P_m(x) = \bigl((1-x^2)P_m^\prime(x) \bigr)^\prime P_n(x) - \bigl((1-x^2)P_n^\prime(x) \bigr)^\prime P_m(x).$$

Therefore,

$$\bigl( n(n+1) - m(m+1) \bigr) \int_{-1}^1 P_n(x) P_m(x) dx$$

$$= \int_{-1}^1 \biggl(\bigl((1-x^2)P_m^\prime(x) \bigr)^\prime P_n(x) - \bigl((1-x^2)P_n^\prime(x) \bigr)^\prime P_m(x)\biggr) dx$$

$$= - \int_{-1}^1 (1-x^2)P_m^\prime(x) P_n^\prime(x) dx + (1-x^2)P_m^\prime(x) P_n(x)\biggr\vert _{x=-1}^{x=1}$$

$$+ \int_{-1}^1 (1-x^2)P_n^\prime(x)P_m^\prime(x) dx + (1-x^2)P_n^\prime(x) P_m(x)\biggr\vert _{x=-1}^{x=1}$$

$$= 0.$$

Since $n \neq m, \; n(n+1) \neq m(m+1)$ and therefore, we have

$$\int_{-1}^1 P_n(x) P_m(x) \; dx = 0 \; {\mbox{ if }} \; m \neq n.$$

height6pt width 6pt depth 0pt

THEOREM 9.4.8   For $n = 0, 1, 2, \ldots$

$$\int_{-1}^1 P_n^2(x) \; dx= \frac{2}{2n+1}.$$ (9.4.11)

Proof. Let us write $V(x) = (x^2 - 1)^n.$ By the Rodrigue's formula, we have

$$\int_{-1}^1 P_n^2(x) \; dx= \int_{-1}^1 \biggl(\frac{1}{n! 2^n} \biggr)^2 \; \frac{d^n}{dx^n} V(x) \frac{d^n}{dx^n} V(x) dx.$$

Let us call $I = \displaystyle\int\limits_{-1}^1 \frac{d^n}{dx^n} V(x) \frac{d^n}{dx^n} V(x) dx.$ Note that for $0 \leq m < n,$

$$\frac{d^m}{dx^m} V(-1) = \frac{d^m}{dx^m} V(1)=0.$$ (9.4.12)

Therefore, integrating $I$ by parts and using (9.4.12) at each step, we get

$$I = \int_{-1}^1 \frac{d^{2n}}{dx^{2n}} V(x) \cdot (-1)^n V(x) dx = (2n)! \int_{-1}^1 (1- x^2)^n dx=(2n)! \; 2 \int_{0}^1 (1- x^2)^n dx.$$

Now substitute $x = \cos \theta$ and use the value of the integral $\int\limits_{0}^{\displaystyle\frac{\pi}{2}} \sin^{2n} \theta \;\;d \theta,$ to get the required result. height6pt width 6pt depth 0pt

We now state an important expansion theorem. The proof is beyond the scope of this book.

THEOREM 9.4.9   Let $f(x)$ be a real valued continuous function defined in $[-1, 1].$ Then

$$f(x) = \sum_{n=0}^\infty a_n P_n(x), \;\; x \in [-1, 1]$$

where $a_n = \displaystyle \frac{2n+1}{2} \int\limits_{-1}^1 f(x) P_n(x) dx.$

Legendre polynomials can also be generated by a suitable function. To do that, we state the following result without proof.

THEOREM 9.4.10   Let $P_n(x)$ be the Legendre polynomial of degree $n.$ Then

$$\frac{1}{\sqrt{1 - 2 x t + t^2}} = \sum_{n=0}^\infty P_n(x) t^n, \;\; t \neq 1.$$ (9.4.13)

The function $h(t) = \displaystyle \frac{1}{\sqrt{1 - 2 x t + t^2}}$ admits a power series expansion in $t$ (for small $t$ ) and the coefficient of $t^n$ in $P_n(x).$ The function $h(t)$ is called the GENERATING FUNCTION for the Legendre polynomials.

EXERCISE 9.4.11  

1. By using the Rodrigue's formula, find $P_0(x), P_1(x)$ and $P_2(x).$
2. Use the generating function (9.4.13)
   1. to find $P_0(x), P_1(x)$ and $P_2(x).$
   2. to show that $P_n(x)$ is an odd function whenever $n$ is odd and is an even function whenever$n$ is even.

Using the generating function (9.4.13), we can establish the following relations:

$$(n+1) P_{n+1}(x) = (2n + 1) \; x \; P_n(x) - n \; P_{n-1}(x)$$ (9.4.14)

$$n P_n(x) = x P_n^\prime(x) - P_{n-1}^\prime(x)$$ (9.4.15)

$$P_{n+1}^\prime(x) = x P_n^\prime(x) + (n+1) P_n(x).$$ (9.4.16)

The relations (9.4.14), (9.4.15) and (9.4.16) are called recurrence relations for the Legendre polynomials, $P_n(x).$ The relation (9.4.14) is also known as Bonnet's recurrence relation. We will now give the proof of (9.4.14) using (9.4.13). The readers are required to proof the other two recurrence relations.

Differentiating the generating function (9.4.13) with respect to $t$ (keeping the variable $x$ fixed), we get

$$-\frac{1}{2} (1 - 2 x t + t^2)^{-\frac{3}{2}} (-2 x + 2 t) = \sum_{n=0}^\infty n P_n(x) t^{n-1}.$$

Or equivalently,

$$(x-t) (1 - 2 x t + t^2)^{-\frac{1}{2}} = (1 - 2 x t + t^2) \sum_{n=0}^\infty n P_n(x) t^{n-1}.$$

We now substitute $\sum\limits_{n=0}^\infty P_n(x) t^n$ in the left hand side for $(1 - 2 x t + t^2)^{-\frac{1}{2}},$ to get

$$(x-t) \sum_{n=0}^\infty P_n(x) t^n = (1 - 2 x t + t^2) \sum_{n=0}^\infty n P_n(x) t^{n-1}.$$

The two sides and power series in $t$ and therefore, comparing the coefficient of $t^n,$ we get

$$x P_n(x) - P_{n-1}(x) = (n+1) P_n(x) + (n-1) P_{n-1}(x) - 2 n \; x \; P_n(x).$$

This is clearly same as (9.4.14).

To prove (9.4.15), one needs to differentiate the generating function with respect to $x$ (keeping $t$ fixed) and doing a similar simplification. Now, use the relations (9.4.14) and (9.4.15) to get the relation (9.4.16). These relations will be helpful in solving the problems given below.

EXERCISE 9.4.12  

1. Find a polynomial solution $y(x)$ of $(1 - x^2) y^{\prime\prime} - 2 x y^\prime + 20 y= 0$ such that $y(1) = 10.$
2. Prove the following:
   1. $\int\limits_{-1}^1 P_m(x) dx = 0$ for all positive integers $m \geq 1.$
   2. $\int\limits_{-1}^1 x^{2n+1} P_{2m}(x) dx = 0$ whenever $m$ and $n$ are positive integers with $m \neq n.$
   3. $\int\limits_{-1}^1 x^{m} P_{n}(x) dx = 0$ whenever $m$ and $n$ are positive integers with $m < n.$
3. Show that $P^\prime_n(1) = \displaystyle \frac{n(n+1)}{2}$ and $P^\prime_n(-1) = (-1)^{n-1}\displaystyle \frac{n(n+1)}{2}.$
4. Establish the following recurrence relations.
   1. $(n+1) P_n(x) = P_{n+1}^\prime(x) - x P_n^\prime(x).$
   2. $(1- x^2) P_n^\prime (x) = n \bigl[ P_{n-1}(x) - x P_n(x) \bigr].$